Levi purchased a total of 50 ride and game tickets and the amusement park and spent $29. if ride tickets cost 75 cents each and

The solution is 20 days, found as follows:
Each of the three friends initially makes 4 bags of dough. After 10 days, each of these bags is split into 4 new bags:
3 friends × 4 bags = 12 bags, and 12 bags × 4 = 48 bags
Then in the following 10 days, 48 bags are again divided into 4 bags each:
48 × 4 = 192 bags
Adding the periods: 10 days + 10 days = 20 days
Therefore, 192 bags are created after 20 days.

Answer:

13.8%

Step-by-step explanation:

With six employees and six checks, there are a total of 36 (6x6) possible distributions. To determine the probability that exactly five individuals receive the correct checks, there’s only one correct distribution for each of them out of the 36 possibilities, resulting in the calculation of 1/36 for the first and then 1/36 for the subsequent checks. Therefore, the final probability becomes 5/36 = 13.8%.

The simplified expression is 10m + 20.5, where m stands for the total number of months. Initially, we can express her expenses as follows: 8.5 + 62.5m and 12 + 3.75m. Adding these two expressions together yields a combined equation of: 10m + 20.5.

Answer:

a) S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) S={top,bottom,side}

c) S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds),(King of diamonds),(Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

Step-by-step explanation:

The sample space consists of all potential outcomes for each experiment:

a) The sample encompasses the outcomes resulting from tossing a coin (heads or tails) and throwing a die (a number between 1 and 6):

S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) A paper cup can settle in one of three different positions: on its top, bottom, or side:

S={top,bottom,side}

c) The entire set of cards within a standard deck constitutes the sample space:

S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds),(King of diamonds),(Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}